Q1. If for ∆ABC and ∆DEF, the correspondence CAB ⟷ EDF gives a congruence, then which of the following is not true?
(a) AC = DE
(b) AB = EF
(c) ∠A = ∠D
(d) ∠B = ∠F
Q2. In ∆ABC and ∆PQR, AB = PQ, AC = PR and ∠B = ∠Q = 90°. Then, ∆ABC ≅ ∆PQR by the criterion
Q4. A polyhedron has 12 faces and 16 vertices. The number of its edges are
Q6. In ∆ABC, AB = 7 cm, BC = 6 cm and CA = 5 cm. Which of the following is correct?
(a) ∠A > ∠B
(b) ∠B > ∠C
(c) ∠C = ∠A
(d) ∠C < ∠A
Q8. Which of the following pairs are supplementary angles?
(a) 54° and 126°
(b) 195° and 126°
(c) 172° and 8°
(d) (1) and (3)
Q9. The sides of a triangle are 6.5 cm, 10 cm and x cm, where x is a positive number. What is the smallest possible value of x among the following?
Q10. Which of the following is correct?
(a) Small “e” and capital “E” are both symmetrical about a line.
(b) Small “o” and capital “O” are both symmetrical about a line.
(c) Small “a” and capital “A” are both symmetrical about a line.
(d) Small “b” and capital “B” are both symmetrical about a line.
Here, ∆ABC ≅ ∆DFE
Therefore, AC = DE, AB = DF and BC = EF
Also, ∠A = ∠D, ∠B = ∠F and ∠C = ∠E
Hence, AB = EF is not correct.
Two right-angled triangles are congruent, by the RHS congruence, if a side and the hypotenuse of one right-angled triangle are equal to the corresponding hypotenuse and side of another right-angled triangle.
Sol. ∠BDC + ∠BDA = 180°
⇒ 119° + ∠BDA = 180°
⇒ ∠BDA = 180° – 119° = 61°
Now, ∠ABD = 180° – (61° + 62°)
= 180° – 123° = 57°
Sol. For any polyhedron that does not intersect itself,
F + V – E = 2
F is the number of faces, V is the number of vertices and E is the number of edges.
12 + 16 – E = 2
⇒ E = 12 + 16 – 2 = 26
Sol. In ∆ABC,
80° + 60° + ∠ACB = 180°
(∵ Sum of three angles of a triangle is 180°)
∠ACB = 180° – 140° = 40°
∠DCE = ∠ACB = 40°
(∵ Vertically opposite angles)
Now in ∆DEC,
40° + 90° + y = 180°
y = 180° – 130° = 50°
We know that the longest side is always opposite the largest interior angle, and the shortest side is always opposite the least interior angle.
Therefore, ∠C > ∠A > ∠B
Hence, ∠A > ∠B
Sol. PQ = TR, PR = TQ, and QR is the common side of two triangles. As all three sides of both triangles are equal, the given triangles are congruent.
The angle corresponding to ∠P is ∠T, and the angle corresponding to ∠Q is ∠R.
Therefore, ∆PQR and ∆TRQ are congruent.
Sol. Two angles are supplementary in they add up to 180 degrees.
54° + 126° = 180°
Also, 172° + 8° = 180°
Sol. We know that the largest side of a triangle is less than the sum of its two other sides.
The least side of a triangle is greater than the difference of its other two sides.
∴ x > 10 – 6.5
⇒ x > 3.5 cm
In the given option, the smallest possible value of x greater than 3.5 is 4.
Sol. Both small “o” and capital “0” are symmetrical about both x-axis and y-axis.